1. If \(x = \cfrac{2\sqrt{15}}{\sqrt{5} + \sqrt{3}}\), then what is the value of \(\cfrac{x + \sqrt{5}}{x - \sqrt{5}} + \cfrac{x + \sqrt{3}}{x - \sqrt{3}}\)?

(a) -1 (b) 0 (c) 1 (d) None of the above

2. সমাধান করো:
\(\cfrac{x+1}{2}+\cfrac{2}{x+1}=\cfrac{x+1}{3}+\cfrac{3}{x+1}-\cfrac{5}{6}\) - translate in english

3. If \(x= \cfrac{2\sqrt{15}}{\sqrt5+\sqrt3}\), then what is the value of \(\cfrac{x+\sqrt5}{x-\sqrt5}\) \(+\cfrac{x+\sqrt3}{x-\sqrt3}\)?

4. If \(x=\cfrac{2\sqrt{15}}{\sqrt5+\sqrt3}\), then determine the value of \(\cfrac{x+\sqrt3}{x-\sqrt3}+\cfrac{x+\sqrt5}{x-\sqrt5}\).

5. If \(\frac{\csc θ + \sin θ}{\csc θ - \sin θ} = \frac{5}{2}\), Let \(\sin θ = x\), then \(\csc θ = \frac{1}{x}\) \[ \frac{\frac{1}{x} + x}{\frac{1}{x} - x} = \frac{5}{2} \Rightarrow \frac{1 + x^2}{1 - x^2} = \frac{5}{2} \Rightarrow 2(1 + x^2) = 5(1 - x^2) \Rightarrow 2 + 2x^2 = 5 - 5x^2 \Rightarrow 7x^2 = 3 \Rightarrow x = \sqrt{\frac{3}{7}} \] ∴ \(\sin θ = \sqrt{\frac{3}{7}}\)

6. \[ \tan 60^\circ = \sqrt{3},\quad \tan 30^\circ = \frac{1}{\sqrt{3}} \\ \cos 60^\circ = \frac{1}{2},\quad \cos 30^\circ = \frac{\sqrt{3}}{2} \\ \sin 60^\circ = \frac{\sqrt{3}}{2},\quad \sin 30^\circ = \frac{1}{2} \] \[ \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \cdot \tan 30^\circ} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{3 - 1}{\sqrt{3}}}{2} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \] \[ \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \] \[ \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{2 + 3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}} \]

7. Let the length of each edge of the cube be \(a\) cm \(\therefore\) The total surface area of the cube = \(6a^2\) square cm If the edge length is increased by 20%, the new edge length = \(a + a \times \cfrac{20}{100}\) cm \(= a + \cfrac{a}{5} = \cfrac{6a}{5}\) cm \(\therefore\) New total surface area of the cube = \(6\left(\cfrac{6a}{5}\right)^2\) square cm = \(\cfrac{216a^2}{25}\) square cm \(\therefore\) Percentage increase in surface area \(= \cfrac{\cfrac{216a^2}{25} - 6a^2}{6a^2} \times 100\%\) \(= \cfrac{(216a^2 - 150a^2) \times 100}{25 \times 6a^2}\%\) \(= \cfrac{4 \times 66a^2}{6a^2}\%\) \(= 44\%\)

8. Translate: \(\cfrac{x}{4 - x} = \cfrac{1}{3x}, \ (x \ne 0, \ x \ne 4)\) — If we express this equation in the form of a quadratic equation \(ax^2 + bx + c = 0\) where \(a \ne 0\), then let's determine the coefficient of \(x\).

(a) 1 (b) 2 (c) 3 (d) 4

9. Solve: \(\frac{x}{x+1} + \frac{x+1}{x} = 2\frac{1}{12}\)

10. \(2x+\cfrac{1}{x}=2\) হলে, \(\cfrac{x}{2x^2+x+1}\) -এর মান কত ? - translate in english

11. Solve: \[ \frac{1}{x} - \frac{1}{3} = \frac{1}{x + 2} - \frac{1}{5} \]

12. If \[ \frac{2x}{3} = \frac{4y}{5} = \frac{7z}{9} \] then find the value of \[ \frac{4x + 12y - 21z}{3y} \]

13. Solve: \[ \frac{x}{x+1} + \frac{x+1}{x} = 2\frac{1}{6} \]

14. Solve: \(\frac{x}{x+1} + \frac{x+1}{x} = 2\frac{1}{2}\)

15. If \(x + \cfrac{1}{x} = 3\), then what is the value of \(\cfrac{(x^2 + 3x + 1)}{(x^2 + 7x + 1)}\)?

(a) \(\cfrac{3}{5}\) (b) \(\cfrac{5}{3}\) (c) \(\cfrac{2}{4}\) (d) \(\cfrac{1}{2}\)

16. If \(x = 2\), \(y = 3\), and \(z = 6\), then calculate the value of \[ \frac{3\sqrt{x}}{\sqrt{y} + \sqrt{z}} - \frac{4\sqrt{y}}{\sqrt{z} + \sqrt{x}} + \frac{\sqrt{z}}{\sqrt{x} + \sqrt{y}} \]

17. Solve: \(\cfrac{1}{x+1} + \cfrac{2}{x+5} = \cfrac{1}{2}\)

(a) \(x \pm 3\) (b) \(x \pm 2\) (c) \(x \pm 5\) (d) \(x \pm 6\)

18. If \(x = 3 + \sqrt{5}\) and \(xy = 4\), determine the value of \(\cfrac{x^2 - xy + y^2}{x^2 + xy + y^2}\).

19. If \(\cfrac{bz+cy}{a} = \cfrac{cx+az}{b} = \cfrac{ay+bx}{c}\), then prove that \(\cfrac{x}{a(b^2+c^2-a^2)} = \cfrac{y}{b(c^2+a^2-b^2)} = \cfrac{z}{c(a^2+b^2-c^2)}\).

20. If the quadratic equation \(ax^2 + 7x + b = 0\) has roots \(\cfrac{2}{3}\) and \(-3\), then find the values of \(a\) and \(b\).

21. If \(\cfrac{x}{y+z} = \cfrac{y}{z+x} = \cfrac{z}{x+y}\), then prove that each ratio is equal to \(\cfrac{1}{2}\) or -1.

22. Let the interest rate be \( r\% \). \(\therefore\) According to the question, \[ 5000\left(1+\cfrac{r}{100}\right)^2 = 5408 \] \[ \left(1+\cfrac{r}{100}\right)^2 = \cfrac{5408}{5000} = \cfrac{676}{625} \] \[ \left(1+\cfrac{r}{100}\right)^2 = \left(\cfrac{26}{25}\right)^2 \] \[ 1+\cfrac{r}{100} = \cfrac{26}{25} \] \[ \cfrac{r}{100} = \cfrac{26}{25} - 1 = \cfrac{1}{25} \] \[ r = \cfrac{100}{25} = 4 \] \(\therefore\) The interest rate is \( 4\% \).

23. If \(\cfrac{x}{y+z}=\cfrac{y}{z+x}=\cfrac{z}{x+y}\), then prove that each ratio is equal to \(\cfrac{1}{2}\) or 1.

24. If \(\cfrac{x}{y+z} = \cfrac{y}{z+x} = \cfrac{z}{x+y}\), then prove that each ratio is equal to \(\cfrac{1}{2}\) or (-1).

25. If \(\cfrac{x}{y} \propto x + y\) and \(\cfrac{y}{x} \propto x - y\), then show that \(x^2 - y^2 =\) constant.

26. If \(\cfrac{x}{y+z}=\cfrac{y}{z+x}=\cfrac{z}{x+y}\), then prove that each ratio is equal to \(\cfrac{1}{2}\) or (-1).

27. If \(\cfrac{x}{y+z}=\cfrac{y}{z+x}=\cfrac{z}{x+y}\), then prove that the value of each ratio is equal to either \(\cfrac{1}{2}\) or \(-1\).

28. \(\cfrac{x}{4 - x} = \cfrac{1}{3x},\ (x ≠ 0,\ x ≠ 4)\) — Let us express this in the form of a quadratic equation \(ax^2 + bx + c = 0\), where \(a ≠ 0\), and determine the coefficient of \(x\).

29. Let us solve: \[ \frac{2}{x^2} - \frac{5}{x} + 2 = 0, \; x \ne 0 \]

30. Let us solve: \[ \frac{x}{x+1} - \frac{x+1}{x} = 2\frac{1}{12}, \; x \ne 0, -1 \]